In the left side: m is an integer, so 2m is an integer. *. . . . . . . . . . . . n is an integer, so 3n is an integer. *
Hence, 2m + 3n (the sum of two integers) is an integer. *

But the right side is an irreducible fraction (or decimal 14.2857...)

We have reached a contradicition, hence our assumption was incorrect.

. . There are no integers m and n that satisfy 14m + 21n .= .101

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*
Reasons: The set of integers is closed under multiplication and addition.

Feb 12th 2007, 11:37 AM

MathStudent1

Thank you Soroban. One question for you. When orginally I posted my thread I made a typo in that 14m + 21n = 100 NOT 101. Does that change the proof at all or does the logic remain the same. Following your proof, am I correct by stating this below:

Prove that there do not exist integers m and n such that: 14m + 21n = 100

Restating the problem - If m and n are integers, then 14m + 21n = 100

Factor out a 7 and we get 7(2m + 3n) = 100
Divide both sides by 7 and we get 2m + 3n = (100/7)
Since m is an integer then 2*m is an integer and since n is an integer then 3*n is and integer.
However, the RHS (100/7) is not an integer and therefore,
2m + 3n (IS NOT EQUAL TO) (100/7)

Hence, we have reached a contradiction of our assumption that m and n are integers.

Therefore, there do not exist integers m and n such that 14m + 21n = 100.

*****Thank you!*****

Feb 12th 2007, 02:38 PM

Soroban

Hello again, MathStudent1!

Yes, the modified proof is still valid . . .. . and you did an excellent job!